20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is ______ × 10^–2 M. (Nearest integer).

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answer is 57.

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Detailed Solution

$\begin{array}{l} M_{F} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}} \\ = \frac{2 \times 20 + 0.5 \times 400}{420} = 0.571 M \\ = 57.1 \times 10^{- 2} M \\ = 57 \end{array}$

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